Question

A copper wire of mass $20g$ carries a current of $6.023A$. What fraction of the total number of free electrons present in the copper wire flows through the cross-section of the copper wire in $2sec$? (gram atomic weight of copper is $63.5g$).

Answer 1

Step 1: Given data and formula:

Given,

$$\begin{gathered} M=63.5g \\ m=20g \\ i=6.023A \\ t=2sec \end{gathered}$$

We know,

$q=it$

Where, $q$ is the charge, $i$ is current and $t$ is the time.

Step 2: Finding the number of electrons in copper wire:

By substituting the values of current and time, we get,

$$\begin{gathered} q=6.023A\times 2s \\ \Rightarrow q=12.046C \end{gathered}$$

Now, the total number of free electrons will be,

$q=ne$

Where $n$ is the number of electrons and $e$ is the charge on a single electron, we get

$n=\frac{12.046C}{1.6\times {10}^{-19}C}=7.53\times {10}^{19}$

Step 3: Calculating the fraction of free electrons flowing in the copper wire:

We know that according to the mole concept,

$mole=\frac{m}{M}=\frac{N_{Cu}}{N_A}$

Where $m$ is the given mass of copper and $M$ is the molar mass of Copper.

$N_{Cu}$ is the number of atoms of Cu.

$N_A$ is Avogadro number.

By substituting all the values we get,

$N_{Cu}=\frac{20}{63.5}\times 6.023\times {10}^{23}$

$N_{Cu}=1.9\times {10}^{23}$

The no. of electrons present in one Cu atom is $29$.

The total no. of electrons would be,

$$\begin{gathered} N=29\times 1.9\times {10}^{23} \\ N=5.5\times {10}^{24} \end{gathered}$$

Now, the fraction of free electrons would be

$$\begin{gathered} \frac{n}{N}=\frac{7.53\times {10}^{19}}{5.5\times {10}^{24}} \\ \frac{n}{N}=1.36\times {10}^{-5} \end{gathered}$$